100 ml of $O_2$ subjected to silent electric discharge till volume of $O_2$ left and $O_3$ formed are equal then

$3O_2\to 2O_3$

 100         0
100-x      $\frac{2x}{3}$
(100-x) =  $\frac{2x}{3}$
X = 60ml

For 100-x, 100-60=40 mL. This means that 60 mL volume of O₂ reacted. 

$\frac{2\times 60}{3}=40 mL$, Volume of $O_3$ formed is 40 ml

Volume of mixture is 80 ml (40 mL unreacted O₂ and 40 mL O₃ formed)

Initial volume is 100 mL and now 80 mL, therefore Decrease in volume 20ml